In the following series of problems, suppose we have a virtual memory system with the following properties: . The memory is byte addressable. . The page size is 1024 bytes. . Physical memory is partitioned into 4 physical pages. • The page table is a single-level design with a total of 8 page table entries. 1. Provide the following properties: a. the number of virtual address bits? b. the number of physical address bits?

In the following series of problems, suppose we have a virtual memory system with the following properties: . The memory is byte addressable. . The page size is 1024 bytes. . Physical memory is partitioned into 4 physical pages. • The page table is a single-level design with a total of 8 page table entries. 1. Provide the following properties: a. the number of virtual address bits? b. the number of physical address bits?

Name: Please Help ASAP!!!! In the following series of problems, suppose we h
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Description: Please Help ASAP!!!! In the following series of problems, suppose we have a virtual memory system with the following properties:• The memory is byte addressable.• The page size is 1024 bytes.• Physical memory is partitioned into 4 physical pages.• Th

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter2: Problem Solving Using C++using

Section2.3: Data Types

Problem 9E: (Practice) Although the total number of bytes varies from computer to computer, memory sizes of...

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(Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GB
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Consider a computer system with a 30-bit logical address and 4-KB page size. The system supports up to 512 MB of physical memory. How many entries are there in each of the following? Assume that each page table entry is 4 Bytes. c. A conventional single-level page table?d. An inverted page table?e. A two-level hierarchical page table? Consider a virtual memory system with a 50-bit logical address and a 38-bit physical address. Suppose that the page/frame size is 16K bytes. Assume that each page table entry is 4 Bytes. a. How many frames are in the systems? How many pages in the virtual address space for a process? b. If a single-level page table is deployed, calculate the size of the page table for each process. c. Design a multilevel page table structure for this system to ensure that each page table can fit into one frame. How many levels do you need? Draw a figure to show your page systems
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7. Consider a logical address space of 256 pages of 1024 words each. This is mapped onto a physical memory of 32 frames. How many bits are there in the logical and physical addresses?
I have a 128-bit computer where addresses are 128 bits wide. I want to implement virtual memory with paging for this computer, and I want to use a page size of 4KB. I will design it so that a page table consumes exactly one page. How many page table entries (PTEs) fit within a page?
Consider a memory system that generates 16-bits addresses and the frame size is 32-byte. At time To the status of the page table is given below. Page Table Entry Page # 1911 11 561 56 644 1068 1884 52 58 42 194 46 a) What is the maximum number of entries within a single page/frame? b) What is the maximum number of page table entries in the memory system? c) Using the given page table, determine the physical addresses for the following logical addresses. Write down the physical addresses in binary and decimal format. (i) (iii) 0101000010001011 1110101110010111 (ii) (iv) 0001100001100111 1110111011101001
Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a multi-level page table design to store all the page table entries of a process, and each page table entry and index entry are 4 bytes in size. What is the total number of page that are required to store the page table entries of a process, across all levels of the page table? You may follow the hint below or finish from scratch to fill the blanks. Please show your calculations to get partial points like 2^10/2^4=2^6. 1. We need to calculate the total number of page table entries needed for a process (i.e., the total number of pages for a process) 2. We need to calculate how many entries each page can store 3. With 1 and 2, we can calculate how many pages needed for the lowest (innermost) level 4. Each page from 3 requires an entry (pointer) in the upper (next) level. We need to calculate how many pages are required to store this next level entries (please note the entry size is always 4…
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. 8 bits are used for offset, 8 bits are used for page # and the max number of pages a process can have is 256. e. Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert hexadecimal numbers to decimal ones.) 0x0389 0xDF78 0x0245 0x8012 f) Now, suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) g)What is the size of the two-level page table…
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Consider a memory-management system based on paging. The total size of the physical address space 64 MB, Pages of size 4 KB, the Logical address space of 4GB. total number of pages are 16384, total number of frames are 16384 and the number of entries in a page table are 1048576.Now Calculate: a)Size of Page Table b) No of bits in Physical Address c) No of Bits in Logical Address
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